3.63.62 \(\int \frac {-12+e^{4+4 x} (-1-x)-15 x-9 x^2-x^3+e^{2+2 x} (7+8 x+6 x^2)+(-x+2 e^{2+2 x} x) \log (x)}{16 x+e^{4+4 x} x+8 x^2+x^3+e^{2+2 x} (-8 x-2 x^2)} \, dx\)

Optimal. Leaf size=29 \[ -x-\log (x)+\frac {2 x+\log (x)}{4-e^{2+2 x}+x} \]

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Rubi [F]  time = 2.15, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {-12+e^{4+4 x} (-1-x)-15 x-9 x^2-x^3+e^{2+2 x} \left (7+8 x+6 x^2\right )+\left (-x+2 e^{2+2 x} x\right ) \log (x)}{16 x+e^{4+4 x} x+8 x^2+x^3+e^{2+2 x} \left (-8 x-2 x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-12 + E^(4 + 4*x)*(-1 - x) - 15*x - 9*x^2 - x^3 + E^(2 + 2*x)*(7 + 8*x + 6*x^2) + (-x + 2*E^(2 + 2*x)*x)*
Log[x])/(16*x + E^(4 + 4*x)*x + 8*x^2 + x^3 + E^(2 + 2*x)*(-8*x - 2*x^2)),x]

[Out]

-x - Log[x] + 7*Log[x]*Defer[Int][(-4 + E^(2 + 2*x) - x)^(-2), x] - 2*Defer[Int][(-4 + E^(2 + 2*x) - x)^(-1),
x] + 2*Log[x]*Defer[Int][(-4 + E^(2 + 2*x) - x)^(-1), x] + 14*Defer[Int][x/(-4 + E^(2 + 2*x) - x)^2, x] + 2*Lo
g[x]*Defer[Int][x/(-4 + E^(2 + 2*x) - x)^2, x] + 4*Defer[Int][x^2/(-4 + E^(2 + 2*x) - x)^2, x] + Defer[Int][1/
(x*(4 - E^(2 + 2*x) + x)), x] - 4*Defer[Int][x/(4 - E^(2 + 2*x) + x), x] - 2*Defer[Int][Defer[Int][(-4 + E^(2
+ 2*x) - x)^(-1), x]/x, x] - 7*Defer[Int][Defer[Int][(4 - E^(2 + 2*x) + x)^(-2), x]/x, x] - 2*Defer[Int][Defer
[Int][x/(4 - E^(2 + 2*x) + x)^2, x]/x, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-12+e^{4+4 x} (-1-x)-15 x-9 x^2-x^3+e^{2+2 x} \left (7+8 x+6 x^2\right )+\left (-x+2 e^{2+2 x} x\right ) \log (x)}{x \left (4-e^{2+2 x}+x\right )^2} \, dx\\ &=\int \left (\frac {-1-x}{x}+\frac {(7+2 x) (2 x+\log (x))}{\left (-4+e^{2+2 x}-x\right )^2}-\frac {-1-2 x+4 x^2+2 x \log (x)}{x \left (4-e^{2+2 x}+x\right )}\right ) \, dx\\ &=\int \frac {-1-x}{x} \, dx+\int \frac {(7+2 x) (2 x+\log (x))}{\left (-4+e^{2+2 x}-x\right )^2} \, dx-\int \frac {-1-2 x+4 x^2+2 x \log (x)}{x \left (4-e^{2+2 x}+x\right )} \, dx\\ &=\int \left (-1-\frac {1}{x}\right ) \, dx-\int \left (\frac {2}{-4+e^{2+2 x}-x}-\frac {1}{x \left (4-e^{2+2 x}+x\right )}+\frac {4 x}{4-e^{2+2 x}+x}-\frac {2 \log (x)}{-4+e^{2+2 x}-x}\right ) \, dx+\int \left (\frac {7 (2 x+\log (x))}{\left (-4+e^{2+2 x}-x\right )^2}+\frac {2 x (2 x+\log (x))}{\left (-4+e^{2+2 x}-x\right )^2}\right ) \, dx\\ &=-x-\log (x)-2 \int \frac {1}{-4+e^{2+2 x}-x} \, dx+2 \int \frac {\log (x)}{-4+e^{2+2 x}-x} \, dx+2 \int \frac {x (2 x+\log (x))}{\left (-4+e^{2+2 x}-x\right )^2} \, dx-4 \int \frac {x}{4-e^{2+2 x}+x} \, dx+7 \int \frac {2 x+\log (x)}{\left (-4+e^{2+2 x}-x\right )^2} \, dx+\int \frac {1}{x \left (4-e^{2+2 x}+x\right )} \, dx\\ &=-x-\log (x)-2 \int \frac {1}{-4+e^{2+2 x}-x} \, dx+2 \int \left (\frac {2 x^2}{\left (-4+e^{2+2 x}-x\right )^2}+\frac {x \log (x)}{\left (-4+e^{2+2 x}-x\right )^2}\right ) \, dx-2 \int \frac {\int \frac {1}{-4+e^{2+2 x}-x} \, dx}{x} \, dx-4 \int \frac {x}{4-e^{2+2 x}+x} \, dx+7 \int \left (\frac {2 x}{\left (-4+e^{2+2 x}-x\right )^2}+\frac {\log (x)}{\left (-4+e^{2+2 x}-x\right )^2}\right ) \, dx+(2 \log (x)) \int \frac {1}{-4+e^{2+2 x}-x} \, dx+\int \frac {1}{x \left (4-e^{2+2 x}+x\right )} \, dx\\ &=-x-\log (x)-2 \int \frac {1}{-4+e^{2+2 x}-x} \, dx+2 \int \frac {x \log (x)}{\left (-4+e^{2+2 x}-x\right )^2} \, dx-2 \int \frac {\int \frac {1}{-4+e^{2+2 x}-x} \, dx}{x} \, dx+4 \int \frac {x^2}{\left (-4+e^{2+2 x}-x\right )^2} \, dx-4 \int \frac {x}{4-e^{2+2 x}+x} \, dx+7 \int \frac {\log (x)}{\left (-4+e^{2+2 x}-x\right )^2} \, dx+14 \int \frac {x}{\left (-4+e^{2+2 x}-x\right )^2} \, dx+(2 \log (x)) \int \frac {1}{-4+e^{2+2 x}-x} \, dx+\int \frac {1}{x \left (4-e^{2+2 x}+x\right )} \, dx\\ &=-x-\log (x)-2 \int \frac {1}{-4+e^{2+2 x}-x} \, dx-2 \int \frac {\int \frac {1}{-4+e^{2+2 x}-x} \, dx}{x} \, dx-2 \int \frac {\int \frac {x}{\left (4-e^{2+2 x}+x\right )^2} \, dx}{x} \, dx+4 \int \frac {x^2}{\left (-4+e^{2+2 x}-x\right )^2} \, dx-4 \int \frac {x}{4-e^{2+2 x}+x} \, dx-7 \int \frac {\int \frac {1}{\left (4-e^{2+2 x}+x\right )^2} \, dx}{x} \, dx+14 \int \frac {x}{\left (-4+e^{2+2 x}-x\right )^2} \, dx+(2 \log (x)) \int \frac {1}{-4+e^{2+2 x}-x} \, dx+(2 \log (x)) \int \frac {x}{\left (-4+e^{2+2 x}-x\right )^2} \, dx+(7 \log (x)) \int \frac {1}{\left (-4+e^{2+2 x}-x\right )^2} \, dx+\int \frac {1}{x \left (4-e^{2+2 x}+x\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.59, size = 43, normalized size = 1.48 \begin {gather*} -x-\frac {2 x}{-4+e^{2+2 x}-x}-\log (x)-\frac {\log (x)}{-4+e^{2+2 x}-x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-12 + E^(4 + 4*x)*(-1 - x) - 15*x - 9*x^2 - x^3 + E^(2 + 2*x)*(7 + 8*x + 6*x^2) + (-x + 2*E^(2 + 2*
x)*x)*Log[x])/(16*x + E^(4 + 4*x)*x + 8*x^2 + x^3 + E^(2 + 2*x)*(-8*x - 2*x^2)),x]

[Out]

-x - (2*x)/(-4 + E^(2 + 2*x) - x) - Log[x] - Log[x]/(-4 + E^(2 + 2*x) - x)

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fricas [A]  time = 0.68, size = 45, normalized size = 1.55 \begin {gather*} -\frac {x^{2} - x e^{\left (2 \, x + 2\right )} + {\left (x - e^{\left (2 \, x + 2\right )} + 3\right )} \log \relax (x) + 2 \, x}{x - e^{\left (2 \, x + 2\right )} + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x*exp(1)^2*exp(x)^2-x)*log(x)+(-x-1)*exp(1)^4*exp(x)^4+(6*x^2+8*x+7)*exp(1)^2*exp(x)^2-x^3-9*x^2
-15*x-12)/(x*exp(1)^4*exp(x)^4+(-2*x^2-8*x)*exp(1)^2*exp(x)^2+x^3+8*x^2+16*x),x, algorithm="fricas")

[Out]

-(x^2 - x*e^(2*x + 2) + (x - e^(2*x + 2) + 3)*log(x) + 2*x)/(x - e^(2*x + 2) + 4)

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giac [A]  time = 1.83, size = 49, normalized size = 1.69 \begin {gather*} -\frac {x^{2} - x e^{\left (2 \, x + 2\right )} + x \log \relax (x) - e^{\left (2 \, x + 2\right )} \log \relax (x) + 2 \, x + 3 \, \log \relax (x)}{x - e^{\left (2 \, x + 2\right )} + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x*exp(1)^2*exp(x)^2-x)*log(x)+(-x-1)*exp(1)^4*exp(x)^4+(6*x^2+8*x+7)*exp(1)^2*exp(x)^2-x^3-9*x^2
-15*x-12)/(x*exp(1)^4*exp(x)^4+(-2*x^2-8*x)*exp(1)^2*exp(x)^2+x^3+8*x^2+16*x),x, algorithm="giac")

[Out]

-(x^2 - x*e^(2*x + 2) + x*log(x) - e^(2*x + 2)*log(x) + 2*x + 3*log(x))/(x - e^(2*x + 2) + 4)

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maple [B]  time = 0.08, size = 69, normalized size = 2.38

method	result	size
risch	\(-\frac {\ln \relax (x )}{{\mathrm e}^{2 x +2}-x -4}-\frac {\ln \relax (x ) {\mathrm e}^{2 x +2}+x \,{\mathrm e}^{2 x +2}-x \ln \relax (x )-x^{2}-4 \ln \relax (x )-2 x}{{\mathrm e}^{2 x +2}-x -4}\)	\(69\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x*exp(1)^2*exp(x)^2-x)*ln(x)+(-x-1)*exp(1)^4*exp(x)^4+(6*x^2+8*x+7)*exp(1)^2*exp(x)^2-x^3-9*x^2-15*x-1
2)/(x*exp(1)^4*exp(x)^4+(-2*x^2-8*x)*exp(1)^2*exp(x)^2+x^3+8*x^2+16*x),x,method=_RETURNVERBOSE)

[Out]

-1/(exp(2*x+2)-x-4)*ln(x)-(ln(x)*exp(2*x+2)+x*exp(2*x+2)-x*ln(x)-x^2-4*ln(x)-2*x)/(exp(2*x+2)-x-4)

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maxima [A]  time = 0.50, size = 40, normalized size = 1.38 \begin {gather*} -\frac {x^{2} - x e^{\left (2 \, x + 2\right )} + 2 \, x - \log \relax (x)}{x - e^{\left (2 \, x + 2\right )} + 4} - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x*exp(1)^2*exp(x)^2-x)*log(x)+(-x-1)*exp(1)^4*exp(x)^4+(6*x^2+8*x+7)*exp(1)^2*exp(x)^2-x^3-9*x^2
-15*x-12)/(x*exp(1)^4*exp(x)^4+(-2*x^2-8*x)*exp(1)^2*exp(x)^2+x^3+8*x^2+16*x),x, algorithm="maxima")

[Out]

-(x^2 - x*e^(2*x + 2) + 2*x - log(x))/(x - e^(2*x + 2) + 4) - log(x)

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mupad [B]  time = 4.55, size = 61, normalized size = 2.10 \begin {gather*} \frac {\ln \relax (x)}{x-{\mathrm {e}}^{2\,x+2}+4}-\ln \relax (x)-x-\frac {2\,\left (2\,x^2+7\,x\right )}{\left ({\mathrm {e}}^{2\,x}-{\mathrm {e}}^{-2}\,\left (x+4\right )\right )\,\left (7\,{\mathrm {e}}^2+2\,x\,{\mathrm {e}}^2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(15*x + log(x)*(x - 2*x*exp(2*x)*exp(2)) + 9*x^2 + x^3 - exp(2*x)*exp(2)*(8*x + 6*x^2 + 7) + exp(4*x)*exp
(4)*(x + 1) + 12)/(16*x + 8*x^2 + x^3 - exp(2*x)*exp(2)*(8*x + 2*x^2) + x*exp(4*x)*exp(4)),x)

[Out]

log(x)/(x - exp(2*x + 2) + 4) - log(x) - x - (2*(7*x + 2*x^2))/((exp(2*x) - exp(-2)*(x + 4))*(7*exp(2) + 2*x*e
xp(2)))

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sympy [A]  time = 0.32, size = 24, normalized size = 0.83 \begin {gather*} - x + \frac {- 2 x - \log {\relax (x )}}{- x + e^{2} e^{2 x} - 4} - \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x*exp(1)**2*exp(x)**2-x)*ln(x)+(-x-1)*exp(1)**4*exp(x)**4+(6*x**2+8*x+7)*exp(1)**2*exp(x)**2-x**
3-9*x**2-15*x-12)/(x*exp(1)**4*exp(x)**4+(-2*x**2-8*x)*exp(1)**2*exp(x)**2+x**3+8*x**2+16*x),x)

[Out]

-x + (-2*x - log(x))/(-x + exp(2)*exp(2*x) - 4) - log(x)

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